Question

For the reversible equilibrium reaction ${K_c} > {K_p}$ at $298K$ and $\Delta H = + 200kJ$, the forward reaction is favoured if:
i.Pressure is increased
ii.Temperature is increased
iii.Temperature is decreased
iv.Increasing the concentration of reactants
A.i, ii, iv
B.i, ii
C.ii, iv
D.i, iii, iv

Answer 1

The relation between ${K_c}$ and ${K_p}$ is,
${K_p} = {K_c}{(RT)^{\Delta n}}$
For reactions having positive value of $\Delta H$ it means that they are endothermic, that is, these reactions consume heat to move forward. While reactions having negative value of $\Delta H$, mean that they are exothermic and evolution of heat takes place.

Complete step by step answer:
${K_c}$ and ${K_p}$ are the equilibrium constants of the reaction, which are related as,
${K_p} = {K_c}{(RT)^{\Delta n}}$
Where $\Delta n =$ number of moles of product $-$ number of moles of reactant
${K_c} > {K_p}$ means $\dfrac{{{K_c}}}{{{K_p}}} > 1$ or $\dfrac{{{K_p}}}{{{K_c}}} < 1$
As, $\dfrac{{{K_p}}}{{{K_c}}} = {(RT)^{\Delta n}}$
So, ${(RT)^{\Delta n}} < 1$
This means the value of $\Delta n$ is less than zero, i.e., it has negative value, which implies that the number of moles of reactant is greater than the number of moles of product.
On increasing the pressure, the side of the reaction (reactant side or the product side) which has greater number of moles tends to convert themselves so as to decrease the number of molecules and hence balance the increase in pressure. So, in this case the reactant will change into the product on an increase in pressure. Thus, favoring forward reaction.
Similarly, on increasing the concentration of the reactants, in order to maintain the equilibrium, the reactants change into the products. Thus, favoring forward reaction.
Now the positive $\Delta H$ value means that the reaction is endothermic. Therefore, on increasing the temperature, i.e., on providing heat the conversion into products is favoured, that is, forward reaction is favoured.

Thus, option A is correct.

Note:
The relation between ${K_c}$ and ${K_p}$ tell a lot about the progress of the reaction.
For example, if ${K_c} > {K_p}$, then $\Delta n < 0$, which means that the number of moles of reactants is greater than the products and product formation is high.
If ${K_c} < {K_p}$, then $\Delta n > 0$, and the product formation is low.
If ${K_c} = {K_p}$, then $\Delta n = 0$, and product formation is optimum.